Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

 

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

思路：队列， 两个变量记录当前level的node数和下一个level的node数。

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<vector<int> > levelOrderBottom(TreeNode *root) {
13         
14         vector<vector<int>> temp=levelOrder(root);
15         vector<vector<int>> res;
16         for(int i=temp.size()-1;i>=0;i--)
17         res.push_back(temp[i]);
18         return res;
19     }
20     
21     vector<vector<int>> levelOrder(TreeNode* root)
22     {
23         vector<vector<int>> res;
24         if(root==NULL)return res;
25         queue<TreeNode*> s;
26         vector<int> level;
27         s.push(root);
28         int level_count=1;//important
29         int next_level_count=0;//important
30         while(!s.empty())
31         {
32             TreeNode* root=s.front();
33             s.pop();
34             level_count--;
35             level.push_back(root->val);
36 
37             if(root->left!=NULL){s.push(root->left);next_level_count++;}
38             if(root->right!=NULL){s.push(root->right);next_level_count++;}
39             
40             if(level_count==0)
41             {
42                 res.push_back(level);
43                 level.clear();
44                 level_count=next_level_count;
45                 next_level_count=0;
46             }
47         }
48         return res;
49     }
50 };

 

posted @ 2014-07-04 12:28  Hicandyman  阅读(149)  评论(0)    收藏  举报